题目链接:http://poj.org/problem?id=3258 River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 15753   Accepted: 6649 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully ju…

River Hopscotch Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 9263 Accepted: 3994 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The e…

River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 15273   Accepted: 6465 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river.…

River Hopscotch Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 21939 Accepted: 9081 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The…

River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 13598   Accepted: 5791 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river.…

River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 6697   Accepted: 2893 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. T…

River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9326   Accepted: 4016 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. T…

https://vjudge.net/problem/POJ-3258 二分最小值,判断需要删去的点的个数,如果大于给定,则直接return 0,则说明该数需要再小. 最后注意,起点是0终点是l,起点可以不加进数组,终点必须加进去!! #include<iostream> #include<cstdio> #include<queue> #include<cstring> #include<algorithm> #include<cmath…

题目链接:http://poj.org/problem?id=3258 题意:给n个石头,起点和终点也是两个石头,去掉这石头中的m个,使得石头间距的最小值最大. 思路:二分石头间的最短距离,每次贪心地check一下是否满足条件即可,具体看代码. AC代码: #include<iostream> #include<stack> #include<vector> #include<algorithm> #include<cmath> using na…

Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and another rock at the end, Lunit…

Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 5473   Accepted: 2379 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement ta…

题目:http://poj.org/problem?id=3258 题意: 一条河长度为 L,河的起点(Start)和终点(End)分别有2块石头,S到E的距离就是L. 河中有n块石头,每块石头到S都有唯一的距离 问现在要移除m块石头(S和E除外),每次移除的是与当前最短距离相关联的石头, 要求移除m块石头后,使得那时的最短距离尽可能大,输出那个最短距离. 和3273差不多... #include <iostream> #include <cstdio> #include <…

/** 大意:给定n个点,删除其中的m个点,其中两点之间距离最小的最大值 思路: 二分最小值的最大值---〉t,若有距离小于t,则可以将前面的节点删除:若节点大于t,则继续往下查看 若删除的节点大于m,说明t,过于大,需要减小:若删除的节点小于m说明t过于小了,t需要增大 **/ #include <iostream> #include <algorithm> using namespace std; ]; int main() { long long l,n,m; cin>…

地址 别人的代码,自己边界总是控制不好,还不知道哪里错了!思维!这种问题代码越简洁反而越不容易错吧.. #include<stdio.h> #include<algorithm> typedef long long ll; using namespace std; ll n,m,L,a[]; bool bi(ll x){ ll i,cnt=,now=; ;i<=n;i++){ if(a[i]-a[now]<=x)cnt++; else now=i; } ; return…

题目:http://poj.org/problem?id=3258 又A一道,睡觉去了.. #include <stdio.h> #include <algorithm> ]; int s, n, m; bool judge(int mid) { , cnt = ; ; i <= n+; i++) { sum += d[i] - d[i-]; if(sum < mid) cnt++; else sum = ; } if(cnt > m) ; ; } int mai…

P2855 [USACO06DEC]河跳房子River Hopscotch 二分+贪心 每次二分最小长度,蓝后检查需要去掉的石子数是否超过限制. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define N 50010 int n,m,L,a[N]; bool check(int lim){ ; ,j=;i<…

POJ3285 River Hopscotch 此题是大白P142页(即POJ2456)的一个变形题,典型的最大化最小值问题. C(x)表示要求的最小距离为X时,此时需要删除的石子.二分枚举X,直到找到最大的X,由于c(x)=m时满足题意,所以最后输出的是ub-1或者lb(lb==ub-1 注意相邻距离小于x的要删除(此处不是小于等于),对于相邻的距离小于x的两个石子,当删除其中一个后,又会产生其他的相邻的石子,直接计数不好计数,不妨用两个标记last,cur,其中last表示上一个石子,cur…

River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9923   Accepted: 4252 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. T…

River Hopscotch 直接中文 Descriptions 每年奶牛们都要举办各种特殊版本的跳房子比赛,包括在河里从一块岩石跳到另一块岩石.这项激动人心的活动在一条长长的笔直河道中进行,在起点和距离起点 L 远的终点各有一块岩石 (1 ≤ L ≤ 10^9).在起点和终点之间,有 N 块岩石 (0 ≤ N ≤ 50000),每块岩石与起点的距离分别为 Di (0 < Di < L). 在比赛过程中,奶牛轮流从起点出发,尝试到达终点,每一步只能从一块岩石跳到另一块岩石.当然,实力不济的奶…

E - River Hopscotch POJ - 3258 Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and…

River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 11031   Accepted: 4737 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river.…

题目传送门 /* 二分:搜索距离,判断时距离小于d的石头拿掉 */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; typedef long long ll; ; const int INF = 0x3f3f3f3f; ll a[MAXN]; int n, m; bool check(ll d) { ; ; ; i&…

River Hopscotch Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 5193 Accepted: 2260 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The e…

River Hopscotch http://poj.org/problem?id=3258 Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 21165   Accepted: 8791 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping…

1650: [Usaco2006 Dec]River Hopscotch 跳石子 Time Limit: 5 Sec  Memory Limit: 64 MB Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes…

http://www.lydsy.com/JudgeOnline/problem.php?id=1650 看到数据和最小最大时一眼就是二分... 但是仔细想想好像判断时不能贪心? 然后看题解还真是贪心..囧. 原来是之前我脑残了. ... 贪心很简单 排序后. 当前点到之前的点的距离<m就累计(相当于删掉这个点,为什么呢?因为这个点假设last到的不是0,那么这个点删了后,因为后边的点的距离大于它,假设后边的点距离减去这个点的距离也是<m,那么显然删去这个点可以得到2个不用删去的点(否则一定要…

Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and another rock at th…

题目真是不好读,大意例如以下(知道题意就非常好解了) 大致题意: 一条河长度为 L,河的起点(Start)和终点(End)分别有2块石头,S到E的距离就是L. 河中有n块石头,每块石头到S都有唯一的距离 问如今要移除m块石头(S和E除外),每次移除的是与当前最短距离相关联的石头,要求移除m块石头后,使得那时的最短距离尽可能大,输出那个最短距离. //Memory Time //420K 391MS #include<iostream> #include<algorithm> usi…

一个不错的二分,注释在代码里 #include <stdio.h> #include <cstring> #include <algorithm> #include <cmath> #include <iostream> using namespace std; ///二分搜索答案,最大化最小值 int main() { int L,n,m; ]; while(~scanf("%d %d %d",&L,&n,&…

<题目链接> 题目大意:现在有起点和终点两个石块,这两个石块之间有N个石块,现在对这N个石块移除M个石块,使得这些石块之间的最短距离最大,注意,起点和终点这两个石块不能被移除. 解题分析: 二分答案典型题,二分最大的最短距离,然后根据这个最短距离对这些石块从左向右进行判断,用一个last记录每一次判断的起点,如果当前石块到last的距离<=最短距离,说明该木块需要被移除.对于最后一个石块,由于最后一个石块不能被移除,所以加入最后一个石块到last的距离<=mid,那么就直接移除la…